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suss Group Title

Xlogy-logz.ylogz-logx.zlogx-logy=1 how to prove this??

  • one year ago
  • one year ago

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  1. suss Group Title
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    it was supposed to x to the power logy-logz y to the power logz-logx z to the power logx-logy

    • one year ago
  2. hartnn Group Title
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    \(\huge x^{\log y-\log z}y^{\log z-\log x}z^{\log x-\log y}\) like this ?

    • one year ago
  3. suss Group Title
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    ah thank god..yes like that

    • one year ago
  4. hartnn Group Title
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    consider \(\huge \log[x^{\log y-\log z}y^{\log z-\log x}z^{\log x-\log y}]\) can u simplify that ?

    • one year ago
  5. hartnn Group Title
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    use, \(\log AB=\log A+\log B \\ \log A^{B}=B\log A\)

    • one year ago
  6. hartnn Group Title
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    you should get that =0 = ln 1

    • one year ago
  7. suss Group Title
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    i dont understand y u multiplied the whole by log????

    • one year ago
  8. hartnn Group Title
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    to be able to simplify

    • one year ago
  9. hartnn Group Title
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    else, nothing can't be really done to that expression

    • one year ago
  10. suss Group Title
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    i have done that problem in this way \[x ^{\log y/z}y ^{\log z/x}z ^{\log x/y}\]

    • one year ago
  11. suss Group Title
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    and i cudnt figure out the next step..... :(

    • one year ago
  12. hartnn Group Title
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    i know, i tried that also....since i could not proceed, i thought of taking log....

    • one year ago
  13. suss Group Title
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    if u dont mind,can u show me the simplification process...??? :)

    • one year ago
  14. hartnn Group Title
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    \(\huge \log[x^{\log y-\log z}y^{\log z-\log x}z^{\log x-\log y}] \\\large =\log[x^{\log y-\log z}]+\log[y^{\log z-\log x}]+\log[z^{\log x-\log y}]\) got this ?

    • one year ago
  15. hartnn Group Title
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    now use \( \\ \log A^{B}=B\log A\)

    • one year ago
  16. hartnn Group Title
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    @suss

    • one year ago
  17. suss Group Title
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    oh sorry man i jus lost connection leeme check it out

    • one year ago
  18. suss Group Title
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    y did u keep the + sign ???

    • one year ago
  19. suss Group Title
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    @hartnn

    • one year ago
  20. hartnn Group Title
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    because \(\log AB=\log A+\log B \\ \)

    • one year ago
  21. suss Group Title
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    ah finally got it .....thx a ton man

    • one year ago
  22. hartnn Group Title
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    sure ? u getting that = 0 ? then write 0=log 1

    • one year ago
  23. suss Group Title
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    yes 100%..i hav to use formula of log ab bak ther ......i really got it man

    • one year ago
  24. hartnn Group Title
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    good :) welcome ^_^

    • one year ago
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